All rings are assumed to be commutative with identity. Let R be a Noetherian integral domain with quotient field K and let alpha be a non-zero element of an algebraic field extension L of K.Also, let pi : R[X] * R[alpha] be the R-algebra homomorphism sending X to alpha and letpsi_<alpha>(X)=X^d+eta_1X^<d-1>+・・・+eta_d be the monic minimal polynomial of alpha over K.Now, put I_<beta>={r*R|rbeta*R}for beta*L,put I_<[alpha]>=I_<eta1>*I_<eta2>*・・・*I_<etad> and put J_<[alpha]>=I_<alpha>c (psi_<alpha> (X)) where c (psi_<alpha> (X)) denotes the R-submodule of K generated by the coefficients of psi_<alpha>(X). Moreover, put I^^-_<[alpha]>=I_<[alpha]>(1, eta_1, ・・・, eta_<d-1>) and put J^^-_<[alpha]>=eta_dI_<[alpha]>. The element alpha is called an anti-integral element of degree d over R if Ker pi=I_<[alpha]>psi_<alpha>(X)R[X]. When alpha is an anti-integral element over R,A=R_<[alpha]> is called an anti-integral extension of R.In case L=K,it its well known that R[alpha] is an anti-integral extension of R if and only if R=R[alpha]*R[alpha^<-1>].Assume that A=R[alpha] is an anti-integral extension of R.Then we obtain the following results :1. I^^-_<alpha>=R if and only if the restriction mapping psi : Spec(A) * Spec(R) is surjective and A is a flat R-module.2. A is an unramified extension over R,that is, the differential module OMEGA_R(A)=(0) if and only if I_<[alpha]>psi'_<alpha>(alpha)A=A.In this case, A is a flat R-module. This means that unramified extensions relate the unit elements.3. A is a flat R-module if and only if I^^-_<[alpha]>A=A.In case L=K,it is equivalent that A is an unramified extension of R.4. Let alpha_i(1<less than or equal>i<less than or equal>n)be elements of L such that the R[alpha_i] are unramified over R.Then R[alpha_1, alpha_2, ・・・, alpha_n] is an unramified extention over R.5.If A is integral over R,then alpha is a unit of A and only if eta_d is a unit of R.